Topic 3 Motion in two dimensions - Gneet 3 Motion in two dimensions e 1 Position of points in two...

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e1

Position of points in two dimensions is represented in vector form

𝑟1 = 𝑥1𝑖̂ + 𝑦1𝑗̂ 𝑎𝑛𝑑 𝑟2 = 𝑥2𝑖̂ + 𝑦2𝑗̂

If particle moves from r1 position to r2 position then

Displacement is final position – initial position

𝑠 = 𝑟2 − 𝑟1

𝑠 = (𝑥2𝑖̂ + 𝑦2𝑗̂) − (𝑥1𝑖̂ + 𝑦1𝑗̂ )

𝑠 = (𝑥2 − 𝑥1)𝑖̂ + (𝑦2 − 𝑦1)𝑗̂

If x2 – x1 = dx and y2 – y1 = dy

𝑠 = 𝑑𝑥𝑖̂ + 𝑑𝑦𝑗̂

Now

�⃗� =𝑑𝑠

𝑑𝑡=

𝑑𝑥

𝑑𝑡𝑖̂ +

𝑑𝑦

𝑑𝑡𝑗̂

dx/dt = vx is velocity of object along x-axis and dy/dt = vy is velocity along y-axis

�⃗� = 𝑣𝑥𝑖̂ + 𝑣𝑦𝑗 ̂

Thus motion in two dimension is resultant of motion in two independent component motions

taking place simultaneously in mutually perpendicular directions.

Magnitude of velocity

𝑣 = √𝑣𝑥2 + 𝑣𝑦

2

Angle between velocity and x-axis is

𝑡𝑎𝑛𝜃 =𝑣𝑦

𝑣𝑥

�⃗� =𝑑𝑣𝑥

𝑑𝑡𝑖̂ +

𝑑𝑣𝑦

𝑑𝑡𝑗̂

�⃗� = 𝑎𝑥𝑖̂ + 𝑎𝑦𝑗̂

Magnitude of acceleration


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e2

𝑎 = √𝑎𝑥2 + 𝑎𝑦

2

Equation of motion in vector form

𝑠 = �⃗⃗�𝑡 +1

2�⃗�𝑡2

�⃗� = �⃗⃗� + �⃗�𝑡

𝑟 = 𝑟0 + �⃗⃗�𝑡 +1

2�⃗�𝑡2

𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =�⃗⃗� + �⃗�

2

𝑣2 − 𝑢2 = 2�⃗� ∙ (𝑟 − 𝑟0)

Solved Numerical

Q1) A particle starts its motion at time t = 0 from the origin with velocity 10j m/s and moves

in X-Y plane with constant acceleration 8i+2j

(a) At what time its x-coordinate becomes 16m? And at that time what will be its

y co-ordinate?

(b) What will be the speed of this particle at this time?

Solution:

From equation of motion

𝑟 = 𝑟0 + �⃗⃗�𝑡 +1

2�⃗�𝑡2

𝑥𝑖̂ + 𝑦𝑗̂ = (10𝑗̂)𝑡 +1

2(8𝑖̂ + 2𝑗̂)𝑡2

𝑥𝑖̂ + 𝑦𝑗̂ = (4𝑡2)𝑖̂ + (10𝑡 + 𝑡2)𝑗̂

Thus

𝑥 = 4𝑡2

When x = 16

16 = 4t2 , t = 2 sec

Now y = 10t+t2 at t = 2 sec


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e3

Y = 20+4 = 240 m

b)

�⃗� = �⃗⃗� + �⃗�𝑡

�⃗� = 10𝑗̂ + (8𝑖̂ + 2𝑗̂)𝑡

t =2 sec

�⃗� = 10𝑗̂ + (8𝑖̂ + 2𝑗̂) × 2

�⃗� = 16𝑖̂ + 14𝑗̂

|𝑣|⃗⃗⃗⃗⃗⃗ = √162 + 142 = 21.26 𝑠𝑒𝑐

Q2) A bomber plane moves due east at 100km/hr over a town X at a certain time, Six minutes

later an interceptor plans sets off from station Y which is 40km due south of X and flies north

east. If both maintain their courses, find the velocity with which interceptor must fly in order

to take over the bomber

Solution:

Let velocity of interceptor speed be V . Now components of

velocity along North direction is Vcos45 and along East direction

is Vsin45

To intercept the bomber , interceptor has to travel a distance of

40 km in t seconds along North direction Thus 40 = Vcos45t

Vt= 40√2 ---- eq(1)

Now total distance travelled by bomber = 100t + distance travelled in 6 minutes before

interceptor takeoff

Distance travelled by bomber along East direction = 100t +10 km

Thus distance travelled by interceptor

100t + 10 = Vsin45t

100t +10 = Vt / √2

From eq(1)

100t +10 =40


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e4

t = 3/10 Hr

Substituting above value of t in equation (1) we get

𝑉 =40√2 × 10

3= 188.56 𝑘𝑚/ℎ𝑟

Q3) A motor boat set out at 11a.m. from a position -6i-2j relative to a marker buoy and

travels at steady speed of magnitude √53 on direct course to intercept a ship. This ship

maintains a steady velocity vector 3i+4j and at 12 noon is at position 3i-4j from the buoy.

Find (a) the velocity vector of the motorboat (b) the time of interception and (c) the position

vector of point of interception from the buoy, if distances are measured in kilometers and

speeds in km/hr

Solution

Let xi+vj be the velocity vector of the motor boat

X2+ y2 = 53 ------eq(1)

Position vector of motor boat after time t = -6i-2j + (xi+yj)t

Since position of ship is given 12 noon to find position of ship at time t, time for ship = (t-1)

Position vector of ship after time t = 3i-j+(3i+4j) ( t-1)

The time of interception is given by

-6i-2j+(xi+yj)t = 3i-j +(3i+4j) ( t-1)

∴-6+xt = 3t

∴xt = 6 +3t -------eq(2)

And -2yt = 4t-5

∴yt = 4t-3 ------eq(3)

From equation (1), (2) and (3)

(6+3t)2 + (4t-3)2 = 53t2

∴36+9t2+36t +16t2-24t+9 = 53t2

∴28t-12t-45 = 0

∴(2t-3) (14t +15) = 0

t = (3/2) hr


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e5

Time of interception = 12.30 pm

Solving equation(2) and (3) x = 7 and y = 2

Velocity vector of motor boat = 7i+2j

Point of interception = -6i-2j + (7i +2j) (3/2) = 9/2 i +j

Relative velocity

Two frame of reference A and B as shown in figure moving with

uniform velocity with respect to each other. Such frame of references

are called inertial frame of reference. Suppose two observers, one in

from A and one from B study the motion of particle P.

Let the position vectors of particle P at some instant of time with

respect to the origin O of frame A be

𝑟𝑃,𝐴 = 𝑂𝑃⃗⃗⃗⃗ ⃗⃗

And that with respect to the origin O’ of frame B be

𝑟𝑃,𝐵 = 𝑂′𝑃⃗⃗⃗⃗ ⃗⃗ ⃗⃗

The position vector of O’ w.r.t O is

𝑟𝐵,𝐴 = 𝑂𝑂′⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

From figure it is clear

𝑂𝑃⃗⃗⃗⃗ ⃗⃗ = 𝑂𝑂′⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ + 𝑂′𝑃⃗⃗⃗⃗ ⃗⃗ ⃗⃗ = 𝑂′𝑃⃗⃗⃗⃗ ⃗⃗ ⃗⃗ + 𝑂𝑂′⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

∴ 𝑟𝑃,𝐴 = 𝑟𝑃,𝐵 + 𝑟𝐵,𝐴

Differentiating above equation with respect to time we get

𝑑

𝑑𝑡(𝑟𝑃,𝐴) =

𝑑

𝑑𝑡(𝑟𝑃,𝐵) +

𝑑

𝑑𝑡(𝑟𝐵,𝐴)

∴ �⃗�𝑃,𝐴 = �⃗�𝑃,𝐵 + �⃗�𝐵,𝐴

Here VP,A is the velocity of the particle w.r.t frame of reference A, VP,B is the velocity of the

particle w.r.t. reference frame B and VB,A is the velocity of frame of reference B with respect

to frame A

Suppose velocities of two particles A and B are respectively VA and VB relative to frame of

reference then velocity (VAB) of A with respect to B is


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e6

�⃗�𝐴𝐵 = �⃗�𝐴 − �⃗�𝐵

And velocity VBA of B relative to A is

�⃗�𝐵𝐴 = �⃗�𝐵 − �⃗�𝐴

Thus

�⃗�𝐴𝐵 = −�⃗�𝐵𝐴

And

|𝑣⃗⃗⃗⃗ 𝐴𝐵| = |�⃗�𝐵𝐴|

|𝑣⃗⃗⃗⃗ 𝐴𝐵| = √𝑉𝐴2 + 𝑉𝐵

2 − 2𝑉𝐴𝑉𝐵𝑐𝑜𝑠𝜃

Also angle α made by the relative velocity with VA is given by

𝑡𝑎𝑛𝛼 =𝑉𝐵𝑠𝑖𝑛𝜃

𝑉𝐴 − 𝑉𝐵𝑐𝑜𝑠𝜃

Solved Numerical

Q4) A boat can move in river water with speed of 8 km/h. This boat has to reach to a place

from one bank of the river to a place which is in perpendicular direction on the other bank of

the river. Then (i) in which direction should the boat has to be moved (ii) If the width of the

river is 600m, then what will be the time taken by the boat to cross the river? The river flows

with velocity 4 km/h

Solution

Suppose the river is flowing in positive X direction as shown in figure .

To reach to a place in the perpendicular direction on the other bank,

the boat has to move in the direction making angle θ with Y direction

as shown in figure. . This angle should be such that the velocity of the

boat relative to the opposite bank is in the direction perpendicular to

the bank . Let VBR = Velocity of boat with respect to river

VBO = Velocity of Boat with respect to Observer on bank

VRO = velocity of river with respect to observer on bank


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e7

�⃗⃗�𝐵𝑂 = �⃗⃗�𝐵𝑅 + �⃗⃗�𝑅𝑂

In vector form

𝑉𝐵𝑂𝑗̂ = −𝑉𝐵𝑅𝑠𝑖𝑛𝜃𝑖̂ + 𝑉𝐵𝑅𝑐𝑜𝑠𝜃𝑗̂ + 𝑉𝑅𝑂𝑖̂

𝑉𝐵𝑂𝑗̂ = −8𝑠𝑖𝑛𝜃𝑖̂ + 8𝑐𝑜𝑠𝜃𝑗̂ + 4𝑖̂

Thus considering only x components

0 = −8𝑠𝑖𝑛𝜃 + 4

⟹θ=30O

𝑉𝐵𝑂 = 8𝑐𝑜𝑠𝜃 , 𝑉𝐵𝑂 = 8𝑐𝑜𝑠30

𝑉𝐵𝑂 = 8√3

2= 4√3 = 6.93 𝑘𝑚/ℎ

Time taken to cross river

𝑡 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦=

0.6

6.93= 0.0865 𝐻𝑟 = 5.2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

Q5) In figure a plane moves due east while pilot points the plane somewhat south of east,

towards a steady wind that blows to the northeast. The plane has

velocity VPW relative to the wind, with an airspeed of 200km/hr directed

at an angle θ south of east. The wind has velocity relative to the

ground, with a speed 40√3 km/hr directed 30O east of north. What is

the magnitude of the velocity of plane VPG relative to the ground, and

what is the value of θ

Solution

�⃗⃗�𝑃𝐺 = �⃗⃗�𝑃𝑊 + �⃗⃗�𝑊𝐺

𝑉𝑃𝐺𝑖̂ = 𝑉𝑃𝑊𝑐𝑜𝑠𝜃𝑖̂ − 𝑉𝑃𝑊𝑠𝑖𝑛𝜃𝑗 + 𝑉𝑊𝐺𝑠𝑖𝑛30𝑖̂ + 𝑉𝑊𝐺𝑐𝑜𝑠30𝑗 ̂

𝑉𝑃𝐺𝑖̂ = 𝑉𝑃𝑊𝑐𝑜𝑠𝜃𝑖̂+𝑉𝑊𝐺𝑠𝑖𝑛30𝑖̂ + 𝑉𝑊𝐺𝑐𝑜𝑠30𝑗̂ − 𝑉𝑃𝑊𝑠𝑖𝑛𝜃𝑗 ̂

Comparing y co-ordinates

0 = 𝑉𝑊𝐺𝑐𝑜𝑠30 − 𝑉𝑃𝑊𝑠𝑖𝑛𝜃

0 = 40√3√3

2− 200𝑠𝑖𝑛𝜃


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e8

𝑠𝑖𝑛𝜃 =60

200=

3

10

⇒ θ =17.45O

Comparing x coordinates

𝑉𝑃𝐺𝑖̂ = 𝑉𝑃𝑊𝑐𝑜𝑠𝜃𝑖+̂𝑉𝑊𝐺𝑠𝑖𝑛30𝑖̂

𝑉𝑃𝐺 = 200𝑐𝑜𝑠17.45 + 40√3𝑠𝑖𝑛30

𝑉𝑃𝐺 = 200𝑐𝑜𝑠17.45 + 40√3𝑠𝑖𝑛30 = 225 𝑘𝑚/ℎ𝑟

Projectile motion

A projectile is a particle, which is given an initial velocity, and then moves under the action of

its weight alone.

When object moves at constant horizontal velocity and constant vertical downward

acceleration, such a two dimension motion is called projectile.

The projectile motion can be treated as the resultant motion of two independent component

motion taking place simultaneously in mutually perpendicular directions. One component is

along the horizontal direction without any acceleration and the other along the vertical

direction with constant acceleration due to gravitational force.

Important terms used in projectile motion

When a particle is projected into air, the angle that the direction of projection makes with

horizontal plane through the point of projection is called the angle of projection, the path,

which the particle describes, is called the trajectory, the distance between the point of

projection and the point where the path meets any plane draws through the point of projection

is its range, the time that elapses in air is called as time of flight and the maximum distance

above the plane during its motion is called as maximum height attained by the projectile

Analytical treatment of projectile motion

Consider a particle projected with a velocity u of an angle

θ with the horizontal earth’s surface. If the earth did not

attract a particle to itself, the particle would describe a

straight line, on account of attraction of earth, however,

the particle describes a curve path


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e9

Let us take origin at the point of projection and x-axis along the surface of earth and

perpendicular to it respectively shown in figure

Here gravitational force is the force acting on the object downwards with constant acceleration

of g downwards. There if no force along horizontal direction hence acceleration along

horizontal direction is zero

Motion in x –direction

Motion in x-direction with uniform velocity

At t = 0, X0 = 0 and ux = ucosθ

Position after time t , x = x0 + ux t

X = (ucosθ) t -----eq(1)

Velocity at t , Vx = ux

Vx= ucosθ ------eq(2)

Motion in y-direction:

Motion in y-direction is motion with uniform acceleration

When t = 0, Y0 = 0, uy = usinθ and ay = -g

After time ‘t’, Vy = yy + ay t

Vy = usinθ – gt -----eq(3)

𝑌 = 𝑌0 + 𝑢𝑦𝑡 +1

2𝑎𝑦𝑡2 − − − 𝑒𝑞(4)

Also,

𝑉𝑦2 = 𝑢𝑦

2 + 2𝑎𝑦𝑡2

𝑉𝑦2 = 𝑢2𝑠𝑖𝑛2𝜃 − 2𝑔𝑦 − − − −𝑒𝑞(5)

Time of flight(T) :

Time of flight is the time duration which particle moves from O to O’.

i.e t = T, Y = 0

From equation (4)


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e10

0 = 𝑢𝑠𝑖𝑛𝜃𝑇 −1

2𝑔𝑇2

𝑇 =2𝑢𝑠𝑖𝑛𝜃

𝑔 − − − −𝑒𝑞(6)

Range of projectile (R):

Range of projectile distance travelled in time T,

i.e. R = Vx × T

R = ucosθT

𝑅 = 𝑢𝑐𝑜𝑠𝜃2𝑢𝑠𝑖𝑛𝜃

𝑔

Since 2sinθcosθ = sin2θ

𝑅 =𝑢2𝑠𝑖𝑛2𝜃

𝑔 − − − −𝑒𝑞(7)

Maximum height reached (H):

At the time particle reaches its maximum height velocity of particle becomes parallel to

horizontal direction i.e. Vy = 0, when Y = H

From equation (5)

0 = 𝑢2𝑠𝑖𝑛2𝜃 − 2𝑔𝐻

𝐻 =𝑢2𝑠𝑖𝑛2𝜃

2𝑔

Equation of trajectory:

The path traced by a particle in motion is called trajectory and it can be known by knowing the

relation between X and Y

From equation (1) and (4) eliminating time t we get

𝑌 = 𝑋𝑡𝑎𝑛𝜃 −1

2

𝑔𝑋2

𝑢2𝑠𝑒𝑐2𝜃

This is trajectory of path and is equation of parabola. So we can say the path of particle is

parabolic


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e11

Velocity and direction of motion after a given time:

After time ‘t’

Vy = ucosθ and Vy = usinθ – gt

Hence resultant velocity

𝑉 = √𝑉𝑋2 + 𝑉𝑌

2

𝑉 = √𝑢2𝑐𝑜𝑠2𝜃 + (𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑡)2

If direction of motion makes an angle α with horizontal

𝑡𝑎𝑛𝛼 =𝑉𝑦

𝑉𝑥

=𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑡

𝑢𝑐𝑜𝑠𝜃

𝛼 = 𝑡𝑎𝑛−1 (𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑡

𝑢𝑐𝑜𝑠𝜃)

Velocity and direction of motion at a given height

At height ‘h’ , Vx = ucosθ

And

𝑉𝑦 = √𝑢2𝑠𝑖𝑛2𝜃 − 2𝑔ℎ

Resultant velocity

𝑉 = √𝑉𝑥2 + 𝑉𝑦

2

𝑉 = √𝑢2 − 2𝑔ℎ

Note that this is the velocity that a particle would have at height h if it is projected vertically

from ground with u

Solved Numerical

Q6) Two particles are projected at the same instant from point A and B on the same horizontal

level where AB = 28 m, the motion taking place in a vertical plane through AB. The particle

from A has an initial velocity of39 m/s at an angle sin-1 (5/13) with AB and the particle from B

has an initial velocity 25 m/s at an angle sin-1(3/5) with BA. Show that the particle would

collide in mid air and find when and where the impact occurs

Solution


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e12

If both the particles have same initial vertical velocity v

then only particles will collide so that for collision both

particles must be at same height

Vertical velocity of particle at A = usinα1 =39(5/13) =15

m/s

Vertical velocity of particle at B = vsinα2 = 25(3/5) = 15 m/s

Since both particles have same vertical velocity they will collide

Since both the particles have travelled in all horizontal distance of 28 m

Therefore 28 = [ucosα1 + vcosα2 ]t

Now sinα1 = 5/13 thus cosα1 = (12/13)

sinα2 = 3/5 thus cosα2 = 4/5

Substituting values in equation (1) we get

28 = [39(12/13) + 25(4/5) ] t

28 = (56) t

t = 0.5 sec. Particles will collide after 0.5 sec

Now Horizontal distance travelled by particle from A =[ ucosα1]t =[ 39 (12/13) ]0.5= 18 m

And vertical distance travelled =[ usinα1]t – (1/2)gt2

=[39(5/13)](0.5) – [0.5(9.8)(0.5)2] = 6.275 m

Hence particle will collide at height of 6.275 m and from 18 m from point A

Q7) A particle is projected with velocity v from a point at ground level. Show that it cannot

clear a wall of height h at a distance x from the point of projection if

𝑣2 ≥ 𝑔(ℎ + √𝑥2 + ℎ2)

Solution:

From the equation for path of projectile

𝑦 = ℎ = 𝑥𝑡𝑎𝑛𝛼 −𝑔𝑥2

2𝑣2

1

𝑐𝑜𝑠2𝛼

ℎ = 𝑥𝑡𝑎𝑛𝛼 −𝑔𝑥2

2𝑣2(1 + 𝑡𝑎𝑛2𝛼)


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e13

This is a quadratic equation in xtanα which is real. Therefore the discriminate cannot be

negative.

𝑔𝑥2

2𝑣2𝑡𝑎𝑛2𝛼 − 𝑥𝑡𝑎𝑛𝛼 +

𝑔𝑥2

2𝑣2+ ℎ = 0

Here

𝑎 =𝑔2

2𝑣2, 𝑏 = 1, 𝑐 =

𝑔𝑥2

2𝑣2+ ℎ

Now Discriminate D = b2-4ac ≥ 0

1 − 4 (𝑔

2𝑣2) (

𝑔𝑥2 + 2𝑣2ℎ

2𝑣2) ≥ 0

1 ≥ 4 (𝑔

2𝑣2) (

𝑔𝑥2 + 2𝑣2ℎ

2𝑣2)

𝑣4 ≥ 𝑔2𝑥2 + 2𝑣2ℎ𝑔

𝑣2 ≥ 𝑔 (ℎ + √𝑥2 + ℎ2)

Q8) Two ships leave a port at the same time. The first steams north-west at 32 km/hr and the

second 40O south of west at 24 km/hr. After what time will they be 160 km apart?

Solution

As show in figure ship goes with 32 km/hr and ship goes with 24 km/hr

We will find relative velocity of Ship A with respect to B so we will get at

what speed ships are going away say VR

�⃗⃗�𝑅 = �⃗⃗�𝐴 − �⃗⃗�𝐵

𝑉𝑅2 = 𝑉𝐴

2 + 𝑉𝐵2 − 2𝑉𝐴𝑉𝐵𝐶𝑜𝑠85

𝑉𝑅 = √1024 + 576 − 2(32)(24)(0.0872)

𝑉𝑅 = 38.3𝑘𝑚

ℎ𝑟

Let t be the time for ships to become 160km apart

38.3×t=160 ∴ t= 4.18 hr

Q9) A object slides off a roof inclined at an angle of 30O to the horizontal for 2.45m. How far

horizontally does it travel in reaching the ground 24.5 m below


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Solution

As shown in figure component of gravitational acceleration along the inclined roof is, a=gsinθ

= 9.8×(1/2) = 4.9 m/s2

v2 = u2 + 2as

v2 = 02 + 2×4.9×2.45 ⟹ v = 4.9m/s

Now horizontal component of velocity which causes horizontal

displacement.

𝑉ℎ = 𝑉𝑐𝑜𝑠30 = 4.9 ×√3

2

Vertical component 𝑉𝑣 = 𝑉𝑠𝑖𝑛30 = 4.9 ×1

2= 2.45 𝑚/𝑠

Vertical displacement ℎ = 𝑢𝑡 +1

2𝑎𝑡2

24.5 = 2.45 × 𝑡 +1

2× 9.8 × 𝑡2

2t2+t-10 = 0

Roots are t = 2.5s and -2s

Neglecting negative time

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 2.5 × 4.9 ×√3

2= 10.6 𝑚

Q10) Two seconds after the projection, a projectile is moving in a direction at 30o to the

horizontal, after one more second, it is moving horizontally. Determine the magnitude and

direction of the initial velocity.

Solution

Given that in two second angle changed to zero from 30o.

Let y component of velocity at angle 30O by v’y and final velocity = zero.

0 = V’y-gt

V’y = g

Now if V’ is the velocity at 30o, then V’y=V’sin30

∴ g= V’ (1/2) or V’ = 2g

And x component which remains unchanged = vcos30

𝑉𝑥 = 2𝑔 ×√3

2= 𝑔√3

Now in 3 seconds after projection velocity along y axis becomes zero

0= Vy – gt but x component remains unchanged


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Vy = 3g

𝑉 = 𝑉𝑥2 + 𝑉𝑦

2

𝑉2 = (𝑔√3)2

+ (3𝑔)2 = 2𝑔√3

Angle 𝑡𝑎𝑛𝜃 =𝑉𝑦

𝑉𝑥=

3𝑔√3

𝑔√3= √3 θ=60O

Q11) A projectile crosses half its maximum height at a certain instant of time and again 10s

later. Calculate the maximum height. If the angle of projection was 30o, calculate the

maximum range of the projectile as well as the horizontal distance it travelled in above 10 s.

Solution

Time taken by projectile to travel from point B to C is 5 sec.

And during this time it travels H/2 distance show by BE. Since point

B is highest point no vertical component of velocity hence it is a

free fall thus

𝐻

2=

1

2× 9.8 × 𝑡2 ∴ H = 245 m

Angle of projection 30o given, from the formula for height

𝐻 =𝑢2𝑠𝑖𝑛2𝜃

2𝑔

245 =𝑢2𝑠𝑖𝑛230

2𝑔=

𝑢2

8𝑔

u = 138.59

From the formula for range

𝑅 =𝑢2𝑠𝑖𝑛2𝜃

2𝑔= 1697.4

In 10s horizontal distance covered by the particle = ucos30 ×10 ≈1200 m

Q12) A projectile is fired into the air from the edge of a 50-m high cliff at an angle of 30 deg

above the horizontal. The projectile hits a target 200 m away from the base of the cliff. What

is the initial speed of the projectile, v?

Solution

Let v be the velocity of projectile

Velocity along x-axis, Vx = vcosθ

Let t be the time of flight

Displacement along x-axis X, 200= (vcosθ) t

𝑡 =200

𝑣𝑐𝑜𝑠𝜃− − − 𝑒𝑞(1)

Velocity along Y-axis vsinθ


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Displacement along Y-axis 50 = −(𝑣𝑠𝑖𝑛𝜃)𝑡 +1

2𝑔𝑡2 − − − 𝑒𝑞(2)

Substituting value of t from eq(1) in eq(2) we get

50 = −(𝑣𝑠𝑖𝑛𝜃)200

𝑣𝑐𝑜𝑠𝜃+

1

2𝑔 (

200

𝑣𝑐𝑜𝑠𝜃)

2

50 = −(𝑡𝑎𝑛𝜃)200 +1

29.8 (

200

𝑣𝑐𝑜𝑠𝜃)

2

50 + (𝑡𝑎𝑛30)200 = 4.9 (200

𝑣𝑐𝑜𝑠30)

2

𝑣 = √4.9 × 2002

𝑐𝑜𝑠230 × [50 + 200𝑡𝑎𝑛30]

V= 39.74 m/s

Q13) A perfectly elastic ball is thrown on wall and bounces back over the head of the thrower,

as shown in the figure. When it leaves the thrower’s hand, the ball is 2m above the ground

and 4m from the wall, and has velocity along x-axis and y-axis is same and is 10m/sec. How

far behind the thrower does the ball hit the ground? [assume g = 10 m/s2]

Solution

Displacement along x axis, 4 = 10(t), ∴ t= 0.4sec

Displacement along y-axis

𝑦 = 10(0.4) −1

210(0.4)2 = 3.2 𝑚

Thus ball struck the wall at height 2+3.2 = 5.2 m

Velocity along y-axis of ball when it struck the wall ( use formula v=u+at) = 10-

10(0.4) = 6m/s

When ball rebounds, direction of y component of velocity remains unchanged. But x

component get’s reversed.

Component of velocity after bounce are vx = -10m/s and vy = 6m/s

Angle with which it rebounds tanθ = 6/10 = 0.6, θ =35o,

With negative x-axis

Now H = 5.2, Time taken to travel H

5.2 = - 6t + ½ ×10×t2

Root of equation t = 1.78 sec.


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Distance travelled along x-axis = 1.78× 10 = 17.8 m

Ball falls behind person = 17.8 – 4 = 13.8 m

Q14) An aeroplane has to go to from a point O to another point A, 500 km away 30o east of

north. A wind is blowing due north at a speed of 20 m/s. the air –speed of the plane is 150

m/s. [ HC Verma, problem 49]

Solution

In fig(a) Vector W represents

direction of velocity of wind.

Vector P represents direction of

plane and vector R represents

position where plane has to go.

In fig b. Wind velocity vector W is

added to velocity vector of plane

P. and R is resultant velocity.

From the geometry of fig(b)

∠OAP = 30o.

By sine formula

150

𝑠𝑖𝑛30=

20

𝑠𝑖𝑛𝛼

Sinα = 0.066 or α = 3o48’

Thus angle between Wind and Plane of velocity is 30+3o 48’ = 33o48’

Thus R2 = W2 +P2 + 2WPcos33o48’

R2 = 202 + 1502 + 2×20×150 (0.8310) =27886

R = 167 m/s

Distance is 500km time taken t

𝑡 =500,000

167= 2994 𝑠

Q15) Airplanes A and B are flying with constant velocity in the same vertical plane at angles

30° and 60° with respect to the horizontal

respectively as shown in figure. The speed of A is

100√3 ms−1 . At time t = 0 s, an observer in A finds B

at distance of 500 m. This observer sees B moving

with a constant velocity perpendicular to the line of

motion of A. If a t = t0, A just escapes being hit by B,

t0 in seconds is

Solution:

From the geometry of figure V = Vatan30

𝑉 = 100√3 ×1

√3= 100𝑚/𝑠

Distance for V is 500 m

to = 500/100 = 5sec

Q16) A rocket is moving in a gravity free space with a constant acceleration of 2 ms−2 along +


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x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from

the left end of the chamber in + x direction with a speed of 0.3 ms−1 relative to the rocket. At

the same time, another ball is thrown in −x direction with a speed of 0.2 ms−1 from its right

end relative to the rocket. The time in seconds when the two balls hit each other is

Solution:

Let A have velocity 0.3m/s and B have velocity 0.2 m/s.

Ball A will face acceleration opposite to its motion as rocket is moving with positive

acceleration Thus A ball will travel first in positive direction then negative direction. Ball B will

travel up to the wall opposite for collision

For ball B direction of acceleration is in the direction of velocity i.e. 2m/s2

4 = 0.2 × 𝑡 +1

2× 2 × 𝑡2

Root is t = 1.99 ≈ 2s

Thus balls will collide after 2sec

Q17) A coin is drop in the elevator from height of 2m. Elevator accelerating down with certain

acceleration. Coin takes 0.8 sec. find acceleration of elevator

Solution:

Since elevator is going down with positive acceleration.

Resultant acceleration of coin = g-a,

2 = 0 +1

2(𝑔 − 𝑎)𝑡2

2 = 0 +1

2(9.8 − 𝑎)(0.8)2

a = 3.55 m/s2

Q18 ) A particle starts its motion at time t = 0 from the origin with velocity 10j m/s and

moves in X-Y plane with constant acceleration 8i+2j

(a) At what time its x-coordinate becomes 16m? And at that time what will be its

y co-ordinate?

(b) What will be the speed of this particle at this time?

Solution:

From equation of motion

𝑟 = 𝑟0 + �⃗⃗�𝑡 +1

2�⃗�𝑡2


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𝑥𝑖̂ + 𝑦𝑗̂ = (10𝑗̂)𝑡 +1

2(8𝑖̂ + 2𝑗̂)𝑡2

𝑥𝑖̂ + 𝑦𝑗̂ = (4𝑡2)𝑖̂ + (10𝑡 + 𝑡2)𝑗̂

Thus

𝑥 = 4𝑡2

When x = 16

16 = 4t2 , t = 2 sec

Now y = 10t+t2 at t = 2 sec

Y = 20+4 = 240 m

b) �⃗� = �⃗⃗� + �⃗�𝑡

�⃗� = 10𝑗̂ + (8𝑖̂ + 2𝑗̂)𝑡

t =2 sec

�⃗� = 10𝑗̂ + (8𝑖̂ + 2𝑗̂) × 2

�⃗� = 16𝑖̂ + 14𝑗̂

|𝑣|⃗⃗⃗⃗⃗⃗ = √162 + 142 = 21.26 𝑠𝑒𝑐

Q19) A particle is projected with certain velocity as shown in figure.

If the time interval required to cross the point B and C is t1

And the time interval required to cross D and E t2

Find t22 – t1

2

Solution : We know that any projectile passes through same height for two time. Velocity of

projectile along Y axis vertical is usinθ

Let point B and C be at height h1. Thus time difference can be calculated as follows by forming

equation of motion

ℎ1 = 𝑢𝑠𝑖𝑛𝜃𝑡 −1

2𝑔𝑡1

2

𝑔𝑡12 − 2𝑢𝑠𝑖𝑛𝜃𝑡 + 2ℎ1 = 0


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Above equation is quadratic equation having two real roots which gives time require to cross

same height while going up which will have smaller value than time required to pass same

height while going down

In this equation a = g , b = 2usinθ and c = 2h1

Discriminate 𝐷 = √𝑏2 − 4𝑎𝑐

On substituting values of a,b c in above equation

√𝐷 = √4𝑢2𝑠𝑖𝑛𝜃2 − 8𝑔ℎ1

If α and β are the two roots then difference in two roots will give us the time difference

between time taken by projectile to cross the same height

𝛼 − 𝛽 =√𝐷

𝑎=

√4𝑢2𝑠𝑖𝑛𝜃2 − 8𝑔ℎ1

𝑔=

2√𝑢2𝑠𝑖𝑛𝜃2 − 2𝑔ℎ1

𝑔

Thus

𝑡1 =2√𝑢2𝑠𝑖𝑛𝜃2 − 2𝑔ℎ1

𝑔

Similarly

𝑡2 =2√𝑢2𝑠𝑖𝑛𝜃2 − 2𝑔ℎ2

𝑔

𝑡22 − 𝑡1

2 =4(𝑢2𝑠𝑖𝑛𝜃2 − 2𝑔ℎ2)

𝑔2−

4(𝑢2𝑠𝑖𝑛𝜃2 − 2𝑔ℎ1)

𝑔2

𝑡22 − 𝑡1

2 =4

𝑔2(2𝑔ℎ1 − 2𝑔ℎ2)

𝑡22 − 𝑡1

2 =8

𝑔(ℎ1 − ℎ2)

Now h1-h2 = h as shown in figure

𝑡22 − 𝑡1

2 =8ℎ

𝑔

Q20) A ball is projected from a point A with velocity 10m/s. Perpendicular to the inclined plane

as shown in figure. Find the range of the ball on inclined plane

Solution:

We will turn the inclined plane and make it horizontal, and solve the

problem in terms of regular projectile problem. Now gravitational acceleration will be, From

figure component of gravitational acceleration perpendicular

to x-axis is gcos30 and acceleration parallel to x-axis will be

gsinθ

When we turn the plane then component of velocity along the

plane will be zero


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Now when projectile reaches the ground displacement along vertical is zero

0 = 10𝑡 −1

2𝑔𝑐𝑜𝑠30𝑡2

𝑡 =40

𝑔√3=

40

10√3=

4

√3

Thus time of flight of projectile is 4/√3

Now gravitation have component along slope is gsin30 as show in figure

Now displacement along the slope = 0 +1/2gsinθt2

𝑅 = 𝑢𝑡 +1

2𝑔𝑠𝑖𝑛30𝑡2

𝑅 = 0 +1

2𝑔𝑠𝑖𝑛30

16

3

If g=10 then R= 40/3 m

Q21) A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is

α =30o and the angle of barrel to the horizontal is β

= 60o. The initial velocity of shell is 21 m/s. Find the

distance from gun to the point at which the shell falls

Solution:

Asked to find OA

Note that projectile fired with angle of 30o with slope.

Now if we rotate the plane then

Velocity along x-axis is ucos30 and velocity along y

axis is usin30

Component of gravitational acceleration perpendicular

to x axis is gcos30 and parallel will be gsin30

This time of flight. Displacement along y-axis is zero

0 = 𝑢𝑠𝑖𝑛30𝑡 −1

2𝑔𝑐𝑜𝑠30𝑡2

𝑡 =2𝑢

𝑔√3

Displacement along x axis Range

𝑅 = 𝑢𝑐𝑜𝑠30𝑡 −1

2𝑔𝑠𝑖𝑛30𝑡2

𝑅 = 𝑢√3

2

2𝑢

𝑔√3−

1

2𝑔

1

2

4𝑢2

3𝑔2


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𝑅 =𝑢2

𝑔−

𝑢2

3𝑔

𝑅 =𝑢2

𝑔(1 −

1

3) =

2𝑢2

3𝑔

𝑅 =2(212)

3 × 9.8= 30 𝑚

Q22) A particle is projected with velocity u strikes at right angle a plane inclined at angle β

with horizontal . find the height of the strike point height

Solution

As show in figure let object strike at point height h,

initial velocity be u and angle of projection be θ

with horizontal

Now gravitational acceleration perpendicular to

inclined to plane.

If we rotate the inclined plane make it horizontal

then angle of projection will be (θ – β)

Component of velocity along x aixs will be ucos(θ –

β) and along y axis is usin (θ – β)

Gravitational acceleration perpendicular to plane is gcosβ and parallel to plane is gsinβ

Particle strike at 90O with horizontal . i.e velocity along x axis is zero, and displacement along

vertical is zero. Thus we will form two equations

At the end of time of flight t velocity along x-axis is zero form equation of motion

V = u – at

0 = 𝑢𝑐𝑜𝑠(𝜃 − 𝛽) − 𝑔𝑠𝑖𝑛𝛽𝑡

𝑡 =𝑢𝑐𝑜𝑠(𝜃 − 𝛽)

𝑔𝑠𝑖𝑛𝛽− − − 𝑒𝑞(1)

Displacement along x-axis That is OA or range

𝑅 = 𝑢𝑐𝑜𝑠(𝜃 − 𝛽)𝑡 −1

2𝑔𝑠𝑖𝑛𝛽𝑡2

Substituting value of t from equation 1 we get

𝑅 = 𝑢𝑐𝑜𝑠(𝜃 − 𝛽)𝑢𝑐𝑜𝑠(𝜃 − 𝛽)

𝑔𝑠𝑖𝑛𝛽−

1

2𝑔𝑠𝑖𝑛𝛽

𝑢2𝑐𝑜𝑠2(𝜃 − 𝛽)

𝑔2𝑠𝑖𝑛2𝛽

𝑅 =𝑢2𝑐𝑜𝑠2(𝜃 − 𝛽)

𝑔𝑠𝑖𝑛𝛽−


2𝑔𝑠𝑖𝑛𝛽

𝑅 =1

2



Since θ is assumed must be eliminated by following equation

Displacement along Y axis zero


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0 = 𝑢𝑠𝑖𝑛(𝜃 − 𝛽)𝑡 −1

2𝑔𝑐𝑜𝑠𝛽𝑡2

𝑡 =2𝑢𝑠𝑖𝑛(𝜃 − 𝛽)


From equation 1 and 3 we get

2𝑢𝑠𝑖𝑛(𝜃 − 𝛽)

𝑔𝑠𝑖𝑛𝛽=

𝑢𝑐𝑜𝑠(𝜃 − 𝛽)

𝑔𝑠𝑖𝑛𝛽

𝑡𝑎𝑛(𝜃 − 𝛽) =

0 = 𝑢𝑐𝑜𝑠(𝜃 − 𝛽) − 2𝑢𝑠𝑖𝑛(𝜃 − 𝛽)

𝑡𝑎𝑛(𝜃 − 𝛽) =1

2𝑐𝑜𝑡𝛽 − − − 𝑒𝑞(4)

From trigonometric identity

𝑠𝑒𝑐2(𝜃 − 𝛽) = 1 + 𝑡𝑎𝑛2(𝜃 − 𝛽)

From equation 4

𝑠𝑒𝑐2(𝜃 − 𝛽) = 1 +1

4𝑐𝑜𝑡2𝛽

As secθ = 1/cosθ thus

𝑐𝑜𝑠2(𝜃 − 𝛽) =4

4 + 𝑐𝑜𝑡2𝛽 − − − 𝑒𝑞(5)

Substituting equation 5 in equation 3 for range we get

𝑅 =1

2

𝑢24

𝑔𝑠𝑖𝑛𝛽(4 + 𝑐𝑜𝑡2𝛽)

𝑅 =2𝑢2

𝑔𝑠𝑖𝑛𝛽(4 + 𝑐𝑜𝑡2𝛽)

Now from figure h = OAsinβ

ℎ =2𝑢2

𝑔𝑠𝑖𝑛𝛽(4 + 𝑐𝑜𝑡2𝛽)𝑠𝑖𝑛𝛽 =

2𝑢2

𝑔(4 + 𝑐𝑜𝑡2𝛽)

Cot may be converted in terms of sine and cos using identity

Q23) To a man walking on straight path at speed of 3km/hr rain appears to fall vertically now

man increased its speed to 6km/hr, it appears that rain is falling at 45o with vertical. Find the

speed of rain Ans 3√2

Solution

It is not given the angle of rain let it be θ with vertical.

Vrm = Velocity of rain with respect to man

Vro = Velocity of man with respect to observer

Vmo= Velocity of man with respect to observer

Now


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Vrm = Vro + Vom or Vrm = Vro – Vmo. Now refer to vector diagram

Case I When speed of man Vmo =3 km/hr

From figure sinθ = Vmo/Vrm = 3/Vrm

Or VrmSiθ = 3 eq(1)

CaseII When speed of man Vmo = 6km/hr

From adjacent figure Resolution of Vro along X-axis and Y

axis are Vrosinθ and VroCosθ

Also note that BA = Vmo - VroSinθ

BA = 6 - Vrosinθ And OA = Vrocosθ

𝑡𝑎𝑛45 =𝐵𝐴

𝑂𝐴=

6 − 𝑉𝑟𝑜𝑠𝑖𝑛𝜃

𝑉𝑟𝑜𝑐𝑜𝑠𝜃

As tan45 = 1 ∴ Vrocosθ = 6 - Vrosinθ

from caseI equation we know that Vrosinθ = 3

Vrocosθ = 6-3 = 3 eq(2) . By squaring and adding

equation 1 and 2 we get Vro =3√2

Q24) Two cars C and D moving in the same direction on a straight road with velocity 12m/s

and 10m/s respectively. When the separation in then was 200m. D starts accelerating so as to

avoid accident.

Solution

Relative velocity of C with respect to D = Vc – VD = 12-10 = 2m/s or u =2m/s

Initial separation or distance is 200m

To avoid accident velocity of D should increased to 12m/s, so that relative velocity of C with

respect to D become zero or v = 0

Relative acceleration of C with respect to D ar = 0 – a = -a

Using formula

v2= u2 +2ar s we get

0 = 4 - 2a(200)

ar = 0.1 m/s


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Q25) A ball rolls off the top of a staircase with a horizontal velocity u m/s. if the steps are h m

high and w m wide the ball will hit the edge of which step.

Solution:

By forming two individual equation in one dimension and by connecting them through time we

will get necessary equation for n

Let ball hit the nth step, there no initial component of velocity along vertical direction, as ball

rolls in horizontal direction

Horizontal displacement =nw eq(1)

Horizontal displacement=ut eq(2)

from eq(1) and eq(2) we get ut=nw

t=nw/u eq(3)

Vertical displacement=nh eq(4)

vertical displacement=½ (g) t2

nh=½ (g) t2 eq(5)

substituting value of time t from eq(3) in eq(5) we get

nh=½ (g) (nw/u)2

𝑛 =2ℎ𝑢2

𝑔𝑤2

Q26) A particle is projected from a point O with velocity u in a direction making an angle α

upward with the horizontal. At P, it is moving at right angle to its initial direction of projection,

What will be velocity at P.

Solution

As shown in figure component of u along x-axis is ucosα

At point P it changes its direction by 90o let velocity be v

Then component of v along x-axis is vcos(90-α) = vsinα

Thus ucosα = vsinα

v = ucotα

Topic 3 Motion in two dimensions - Gneet 3 Motion in two dimensions e 1 Position of points in two...

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